In an explosion a body breaks up into pieces of unequal masses. In this case

Question

In an explosion a body breaks up into pieces of unequal masses. In this case (A) both parts will have numerically equal momentum (B) lighter part will have more momentum (C) heavier part will have more momentum (D) both parts will have equal kinetic energy

Tardigrade Admin · Accepted Answer

From law of conservation of linear momentum
i.e., final momentum = initial momentum

m_{q} v_{1} + m_{2} v_{2} = 0

...(i)
where

m_{1}

and

m_{2}

are masses of pieces.
(initial momentum is zero because body is stationary)
From Eq. (i),

m_{1}, v_{1} = - m_{2} v_{2}

So, numerical value of momentum of both pieces will be equal.

Q. In an explosion a body breaks up into pieces of unequal masses. In this case

both parts will have numerically equal momentum

lighter part will have more momentum

heavier part will have more momentum

both parts will have equal kinetic energy